Question

Balance the following equations: a. \(\operatorname{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) \(\operatorname{Pb}(\mathrm{OH})_{2}(s)+\mathrm{NaNO}_{3}(a q)\) b. \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{Al}\left(\mathrm{OH}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\right.\) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

a. \[\text{Pb}(\text{NO}_3)_2(aq) + 2\text{NaOH}(aq) \rightarrow \text{Pb}(\text{OH})_2(s) + 2\text{NaNO}_3(aq)\] b. \[\text{C}_{12}\text{H}_{22}\text{O}_{11}(l) + 12\text{O}_2(g) \rightarrow 12\text{CO}_2(g) + 11\text{H}_2\text{O}(l)\] c. \[2\text{Al}(\text{OH})_3(s) + 3\text{H}_2\text{SO}_4(aq) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + 6\text{H}_2\text{O}(l)\]

Step-by-step solution

Write the Unbalanced Equation for a

Start with the given unbalanced equation: \[\text{Pb}(\text{NO}_3)_2(aq) + \text{NaOH}(aq) \rightarrow \text{Pb}(\text{OH})_2(s) + \text{NaNO}_3(aq)\]

Balance Lead (Pb) and Sodium (Na)

There is 1 Pb atom on both sides of the equation and 1 Na atom on both sides. So Pb and Na are balanced.

Balance Nitrate (NO\textsubscript{3}) and Hydroxide (OH)

We have 2 NO\textsubscript{3} groups on the left and only 1 on the right. We also have 2 OH groups on the right after balancing NO\textsubscript{3}. Adjust the coefficients: \[\text{Pb}(\text{NO}_3)_2(aq) + 2\text{NaOH}(aq) \rightarrow \text{Pb}(\text{OH})_2(s) + 2\text{NaNO}_3(aq)\]

Write the Unbalanced Equation for b

Start with the given unbalanced equation: \[\text{C}_{12}\text{H}_{22}\text{O}_{11}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)\]

Balance Carbon (C)

We have 12 C on the left and 1 on the right. Add the coefficient 12 to CO_2 on the right: \[\text{C}_{12}\text{H}_{22}\text{O}_{11}(l) + \text{O}_2(g) \rightarrow 12\text{CO}_2(g) + \text{H}_2\text{O}(l)\]

Balance Hydrogen (H)

We have 22 H on the left and 2 on the right. Add the coefficient 11 to H_2O on the right: \[\text{C}_{12}\text{H}_{22}\text{O}_{11}(l) + \text{O}_2(g) \rightarrow 12\text{CO}_2(g) + 11\text{H}_2\text{O}(l)\]

Balance Oxygen (O)

On the left, we have 11 O from C12H22O11 and unknown from O2. On the right, we have 24 O from CO2 and 11 O from H2O. Total on right is 35. We need 12 more O from O2 on the left. Add the coefficient 12: \[\text{C}_{12}\text{H}_{22}\text{O}_{11}(l) + 12\text{O}_2(g) \rightarrow 12\text{CO}_2(g) + 11\text{H}_2\text{O}(l)\]

Write the Unbalanced Equation for c

Start with the given unbalanced equation: \[\text{Al}(\text{OH})_3(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + \text{H}_2\text{O}(l)\]

Balance Aluminum (Al) and Sulfate (SO\textsubscript{4})

On the left, we have 1 Al and 1 SO_4; on the right, 2 Al and 3 SO_4. Add coefficients to adjust: \[2\text{Al}(\text{OH})_3(s) + 3\text{H}_2\text{SO}_4(aq) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + \text{H}_2\text{O}(l)\]

Balance Oxygen (O) and Hydrogen (H)

Adjust the coefficient for water on the right if needed, checking all atoms: \[2\text{Al}(\text{OH})_3(s) + 3\text{H}_2\text{SO}_4(aq) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + 6\text{H}_2\text{O}(l)\]

Key concepts

stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative aspects of chemical reactions. It allows us to determine the amounts of reactants and products involved in a chemical reaction. By using balanced chemical equations and mole ratios, we can predict how much of each substance is required or produced.
For example, in the equation \(\text{Pb}(\text{NO}_3)_2 + 2\text{NaOH} \rightarrow \text{Pb}(\text{OH})_2 + 2\text{NaNO}_3\), stoichiometry tells us that 1 mole of lead(II) nitrate reacts with 2 moles of sodium hydroxide to produce 1 mole of lead(II) hydroxide and 2 moles of sodium nitrate.
Understanding stoichiometry helps in various applications such as calculating yields, optimizing reactions, and even in everyday tasks like baking, where precise ingredient amounts are critical.

chemical reactions

Chemical reactions involve the transformation of reactants into products via breaking and forming chemical bonds. These activities are central to chemistry and are represented by chemical equations.
In our example reactions:

• Lead Nitrate reacts with Sodium Hydroxide: \(\text{Pb}(\text{NO}_3)_2(aq) + 2\text{NaOH}(aq) \rightarrow \text{Pb}(\text{OH})_2(s) + 2\text{NaNO}_3(aq)\)
• Sucrose reacts with Oxygen: \(\text{C}_{12}\text{H}_{22}\text{O}_{11}(l) + 12\text{O}_2(g) \rightarrow 12\text{CO}_2(g) + 11\text{H}_2\text{O}(l)\)
• Aluminum Hydroxide reacts with Sulfuric Acid: \(2\text{Al}(\text{OH})_3(s) + 3\text{H}_2\text{SO}_4(aq) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + 6\text{H}_2\text{O}(l)\)

Chemical reactions can be classified in various ways:

• Combustion
• Synthesis
• Decomposition
• Single replacement
• Double replacement.

In our case, we deal with double replacement and combustion reactions, showcasing different reaction types and mechanisms.

conservation of mass

The law of conservation of mass states that mass is neither created nor destroyed in a closed system through a chemical reaction. This principle is crucial for balancing chemical equations. Each atom must be accounted for on both sides of the equation.
For example:

• Pb is balanced: \(\text{Pb}(\text{NO}_3)_2 + 2\text{NaOH} \rightarrow \text{Pb}(\text{OH})_2 + 2\text{NaNO}_3\)
• C is balanced: \(\text{C}_{12}\text{H}_{22}\text{O}_{11} + 12\text{O}_2 \rightarrow 12\text{CO}_2 + 11\text{H}_2\text{O}\)

By balancing these equations, we respect the conservation of mass. It's like balancing a scale; what's on one side must equally match the other side.
Understanding this concept ensures that chemical equations are correctly balanced, ensuring accurate calculations in stoichiometry and predicting reaction outcomes.

chemical equations

Chemical equations are a symbolic representation of chemical reactions. They show the reactants transforming into products with the use of chemical symbols and formulas.
Writing and balancing chemical equations involves:

• Writing the unbalanced equation with correct chemical formulas
• Identifying and counting all atoms on both sides
• Adjusting coefficients to balance each element

For instance:
1. Write: \(\text{Al}(\text{OH})_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}\)
2. Balance Al and SO\textsubscript{4}: \(2\text{Al}(\text{OH})_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O}\)
3. Balance H and O: \(2\text{Al}(\text{OH})_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}\)
Chemical equations are the language of chemistry, providing a concise way to convey how substances transform during chemical reactions. They are essential for communicating and understanding chemical processes efficiently.